Optimal. Leaf size=153 \[ -\frac{\left (6 a^2+1\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{a+b x}\right )}{6 b^3}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{5 a \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{6 b^3}-\frac{x \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{6 b^2}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x) \]
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Rubi [A] time = 0.0985505, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6321, 5468, 3782, 3770, 3767, 8} \[ -\frac{\left (6 a^2+1\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{a+b x}\right )}{6 b^3}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{5 a \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{6 b^3}-\frac{x \sqrt{\frac{-a-b x+1}{a+b x+1}} (a+b x+1)}{6 b^2}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x) \]
Antiderivative was successfully verified.
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Rule 6321
Rule 5468
Rule 3782
Rule 3770
Rule 3767
Rule 8
Rubi steps
\begin{align*} \int x^2 \text{sech}^{-1}(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x \text{sech}(x) (-a+\text{sech}(x))^2 \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{b^3}\\ &=\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int (-a+\text{sech}(x))^3 \, dx,x,\text{sech}^{-1}(a+b x)\right )}{3 b^3}\\ &=-\frac{x \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^2}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int \left (-2 a^3+\left (1+6 a^2\right ) \text{sech}(x)-5 a \text{sech}^2(x)\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{6 b^3}\\ &=-\frac{x \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^2}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)+\frac{(5 a) \operatorname{Subst}\left (\int \text{sech}^2(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{6 b^3}-\frac{\left (1+6 a^2\right ) \operatorname{Subst}\left (\int \text{sech}(x) \, dx,x,\text{sech}^{-1}(a+b x)\right )}{6 b^3}\\ &=-\frac{x \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^2}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)-\frac{\left (1+6 a^2\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{a+b x}\right )}{6 b^3}+\frac{(5 i a) \operatorname{Subst}\left (\int 1 \, dx,x,-i \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)\right )}{6 b^3}\\ &=\frac{5 a \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^3}-\frac{x \sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{6 b^2}+\frac{a^3 \text{sech}^{-1}(a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{sech}^{-1}(a+b x)-\frac{\left (1+6 a^2\right ) \tan ^{-1}\left (\frac{\sqrt{\frac{1-a-b x}{1+a+b x}} (1+a+b x)}{a+b x}\right )}{6 b^3}\\ \end{align*}
Mathematica [C] time = 0.258412, size = 200, normalized size = 1.31 \[ \frac{\sqrt{-\frac{a+b x-1}{a+b x+1}} \left (5 a^2+a (4 b x+5)-b x (b x+1)\right )-2 a^3 \log (a+b x)+2 a^3 \log \left (a \sqrt{-\frac{a+b x-1}{a+b x+1}}+b x \sqrt{-\frac{a+b x-1}{a+b x+1}}+\sqrt{-\frac{a+b x-1}{a+b x+1}}+1\right )+i \left (6 a^2+1\right ) \log \left (2 \sqrt{-\frac{a+b x-1}{a+b x+1}} (a+b x+1)-2 i (a+b x)\right )+2 b^3 x^3 \text{sech}^{-1}(a+b x)}{6 b^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.244, size = 189, normalized size = 1.2 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{ \left ( bx+a \right ) ^{3}{\rm arcsech} \left (bx+a\right )}{3}}-{\rm arcsech} \left (bx+a\right ) \left ( bx+a \right ) ^{2}a+{\rm arcsech} \left (bx+a\right ) \left ( bx+a \right ){a}^{2}-{\frac{{\rm arcsech} \left (bx+a\right ){a}^{3}}{3}}+{\frac{bx+a}{6}\sqrt{-{\frac{bx+a-1}{bx+a}}}\sqrt{{\frac{bx+a+1}{bx+a}}} \left ( 2\,{a}^{3}{\it Artanh} \left ({\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}} \right ) +6\,{a}^{2}\arcsin \left ( bx+a \right ) - \left ( bx+a \right ) \sqrt{1- \left ( bx+a \right ) ^{2}}+6\,a\sqrt{1- \left ( bx+a \right ) ^{2}}+\arcsin \left ( bx+a \right ) \right ){\frac{1}{\sqrt{1- \left ( bx+a \right ) ^{2}}}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{2 \, b^{3} x^{3} \log \left (\sqrt{b x + a + 1} \sqrt{-b x - a + 1} b x + \sqrt{b x + a + 1} \sqrt{-b x - a + 1} a + b x + a\right ) - 2 \, b^{3} x^{3} \log \left (b x + a\right ) - 2 \, b x +{\left (a^{3} + 3 \, a^{2} + 3 \, a + 1\right )} \log \left (b x + a + 1\right ) - 2 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (b x + a\right ) +{\left (a^{3} - 3 \, a^{2} + 3 \, a - 1\right )} \log \left (-b x - a + 1\right )}{6 \, b^{3}} + \int \frac{b^{2} x^{4} + a b x^{3}}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} e^{\left (\frac{1}{2} \, \log \left (b x + a + 1\right ) + \frac{1}{2} \, \log \left (-b x - a + 1\right )\right )} - 1\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.35117, size = 720, normalized size = 4.71 \begin{align*} \frac{2 \, b^{3} x^{3} \log \left (\frac{{\left (b x + a\right )} \sqrt{-\frac{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right ) + a^{3} \log \left (\frac{{\left (b x + a\right )} \sqrt{-\frac{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{x}\right ) - a^{3} \log \left (\frac{{\left (b x + a\right )} \sqrt{-\frac{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - 1}{x}\right ) -{\left (6 \, a^{2} + 1\right )} \arctan \left (\frac{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt{-\frac{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) -{\left (b^{2} x^{2} - 4 \, a b x - 5 \, a^{2}\right )} \sqrt{-\frac{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{6 \, b^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{asech}{\left (a + b x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arsech}\left (b x + a\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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